Question

if x^_ is the arithmetic mean of n observation x1,x2,.......,xn, then find the arithmetic mean of x1+a,x2+a,xn+a

Answer 1

Answer: 0

Step-by-step explanation:

As x is aritmetic mean...

x = (x1+x2+........xn)/n

=> x1+x2+........xn = x*n

given (x1-x)+(x2-x)+........+(xn-x)

.=> (x1+x2+.....+xn) - (x+x+x+......n times)

=> ( nx)-(nx)

. = 0......

Hope it will help....☺

Answer 2

Given that :-

$x(bar) = \frac{x_{1}+x_{2}+x_{3}...+x_{n}}{n}$

So,

From here we get :-

$x(bar) n= x_{1}+x_{2}+x_{3}...+x_{n}$

$\rule{200}{2}$

Now,

New observation :-

$[x_{1}+a] +[x_{2}+a] +[x_{3}+a] ...+[x_{n}+a]$

The means of this observation will be as follows :-

$= \frac{[x_{1}+a] +[x_{2}+a] +[x_{3}+a] ...+[x_{n}+a]}{n} \\ \\ \\ = \frac{x_{1}+x_{2}+x_{3}...+x_{n} + an} {n} \\ \\ \\ = \frac{x(bar)n+an} {n} \\ \\ \\ = x(bar) +a$

$\rule{200}{4}$

So,

New arithmetic mean will be equal to

$x(bar) + a$